I am trying to show that the square of any Fibonacci Number differs by one from the product of its neighbors:

$\displaystyle (F_n)^2 - ((F_{n-1}) \times (F_{n+1})) = \pm 1 $

My best attempt is to substitute Binet's Formula:

$\displaystyle \left( \frac { ( 1 + \sqrt{5} )^n - (1 - \sqrt{5} )^n } { 2^n \sqrt{5} } \right)^2 - \left( \frac { ( 1 + \sqrt{5} )^{n-1} - (1 - \sqrt{5} )^{n-1} } { 2^{n-1} \sqrt{5} } \times \frac { ( 1 + \sqrt{5} )^{n+1} - (1 - \sqrt{5} )^{n+1} } { 2^{n+1} \sqrt{5} } \right) $

and simplify, but keep getting buggered near the end. I get down to:

$\displaystyle \frac { ( ( 1 + \sqrt{5} )^{n-1} ( 1 - \sqrt{5} )^{n+1} ) + ( ( 1 - \sqrt{5} )^{n-1} ( 1 + \sqrt{5} )^{n+1} ) - 2 ( -4 )^n } { 5 ( 4^n ) } $

and can go no further due to the unequal bases. I know (because even values of "n" yield -1) that:

$\displaystyle { ( ( 1 + \sqrt{5} )^{n-1} ( 1 - \sqrt{5} )^{n+1} ) + ( ( 1 - \sqrt{5} )^{n-1} ( 1 + \sqrt{5} )^{n+1} ) = - 3 ( -4 )^n } $

which would make the final formula:

$\displaystyle \frac { - 5 ( -4 )^n } { 5 ( 4^n ) } = \pm 1 $

but I seem to be missing something. I could swear I've seen this done, but can't find it anywhere. Any pointers as to what I'm missing?

- Stephen